# Tractor Sled Pulling

1. Aug 7, 2007

### CompCobra

I need a little help understanding something. I was wondering if it is possible to understand the physics behind sled pulling? From what I can tell is there are two major factors in pulling 1) being the coefficient of friction needed to move the sled (14,000lbs of weight over 2, 5'x1' pieces of rubber,) 2) the wheelspeed of the tractor needed to maintain that coefficient of friction (the greater the wheelspeed the more the tire melts and grip is lost.)

Is something like this possible to figure out? Has it been looked at before? What other information do I need to gather or supply to determine the solution?

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2. Aug 7, 2007

### ice109

you haven't defined a problem? there's a lot of physics involved anything. you have to decide what you want to know about tractor pulling before we can tell you about it or at least tell you how difficult it is to figure out.

3. Aug 7, 2007

### CompCobra

Sorry, I would like to know the maximum wheel speed limit to move the sled.

If the wheel speed is too great the rubber melts and looses traction. If the wheel speed is too slow there isn't enough momentum to keep the sled moving as the resistance increases.

4. Aug 7, 2007

### mgb_phys

I don't think that there is a minimum speed. The sled starts off stationary so there is no momentum to keep, the idea is probably only to go as fast as possible because it is a race. The upper limit is as you say the tires and probably the torque curve of the engine.
If you wished to move the sleds economically you would actaully use the slowest speed possible just to keep them moving and overcome static friction.

5. Aug 7, 2007

### rcgldr

This depends on the static and dynamic friction factors of the tire used for the pull. Ideally, the clutch should be doing all the slipping. Back around 1970 there was a time where 1/4 ET for fuel dragsters was about the same whether their tires spun or not on a run. Since then, clutch and tire technology have improved significantly. However in a contest like a tractor pull, I'm not sure if there are specified tires to use to make the field equal.

Normally I see tractor pulls on the dirt, where some slippage is optimal.

6. Aug 7, 2007

### CompCobra

Everyone in a class has to run the same size tire. You can very the rubber compounds either hard or soft. The sled does start out stationary and the object is to pull it 300ft. The momentum build up is used to overcome some of the friction as the weight of the sled is transferred forward onto the skid increasing the friction.

What is required to move 14000lbs sliding on rubber over asphalt with a 1700lb tractor? I can calculate the force required to overcome sliding friction on the sled. Knowing that I should be able to determine an angular velocity for the tractors tire, right? Then if I know the radius of the tire I can get my RPM's. Am I on the right track with that thought process?

Last edited: Aug 7, 2007
7. Aug 8, 2007

### ice109

how can you figure out the angular velocity of the wheels from the force?

you can figure out the maximum torque you can apply to the wheels without them slipping and from that you can figure out your rpms using a common formula that floats around ( i don't remember it ). what is your objective? what exactly are you trying to figure out?

the questions "what is required to move 14000lbs with a 1700lb tractor?" is still too vague. just tell us what you're trying to do? are you trying to apply physics to help you win a tractor race? are you trying to model a tractor race?

if you're trying to figure out how many rpms you should be looking for on your tach while pulling i don't think physics will help you. you're better off talking to someone with a lot of practice and practicing yourself. as a physics problem it has too many degrees of freedom; there's too much involved.

8. Aug 8, 2007

### CompCobra

I was reaching there on the angular velocity.

But essentially I am trying to apply physics to help win a race.

9. Aug 8, 2007

### ice109

that's like trying to apply physics to winning a basketball game. it's pointless. your best bet is to talk to someone with experience in the sport. or any engineer that might be able to help you optimize your setup.

10. Aug 8, 2007

### rcgldr

The obvious things to use are a very heavy flywheel to build up energy at the start of a pull, then utilizing this energy during the last part of a pull. If it's allowed, a mechanically programmable clutch as used on fuel dragsters that keeps the tires near the limits of traction on a run. These clutches use weights, springs, and leverage on both the driving and driven plates to control clutch pressure.

11. Aug 8, 2007

### pervect

Staff Emeritus
This makes sense, it boils down to a choice of a tire with a high coefficient of friction.

Sorry, you've totally lost me with that remark.

The wheelspeed of the tractor will be determined entirely by the velocity at which it is moving and the size of the wheel. I don't see how it will have anything to do with the coefficient of friction.

From a physics point of view, what is needed is that the engine plus the geartrain can supply a constant torque, torque = force * radius of wheel as you increase the speed of the sled.

The faster the sled moves, the more horesepower an engine must have to provide the torque.

The RPM of the engine will be determined by the gearing, and the power of the engine will be a function of the RPM. Thus we have a desired maximumm torque - the engine will need to have enough power to generate this torque at a higher rpm (horsepower is proportional to torque * rpm). Note that in general the power output of an engine will depend on its rpm, matching the two is a job for the gearing system.

Bottom line:

you need good tires (high coefficient of friction) a good engine (high constant torque determined by how much the tires can transfer, the constant torque requires higher horsepower the faster the engine goes), and a gearing system suited to the engine.

Last edited: Aug 8, 2007
12. Aug 8, 2007

### mgb_phys

I think part of the stratergy is to deliberately spin the wheels and partly melt the tire so increasing the coefficient of friction.